Leo Giugiuc's Cyclic Inequality with a Constraint

Solution 1

Homogenize the inequality as

\displaystyle \begin{align} &\frac{a^3+b^3+c^3}{abc}+\frac{16abc}{(a+b)(b_c)(c+a)}\ge 5&\Leftrightarrow\\ &\frac{a^3+b^3+c^3+3abc}{abc}+\frac{16abc}{(a+b)(b+c)(c+a)}\ge 8&\Leftrightarrow\\ &\frac{a^3+b^3+c^3+3abc}{2abc}+\frac{8abc}{(a+b)(b+c)(c+a)}\ge 4&\Leftrightarrow \end{align}

Set $\displaystyle \frac{2a}{b+c}=x\gt 0,$ $\displaystyle \frac{2b}{c+a}=y\gt 0,$ $\displaystyle \frac{2c}{a+b}=z\gt 0.$ Then $\displaystyle xy+yx+zx+\frac{8abc}{(a+b)(b+c)(c+a)}=4.$ So the latter inequality is equivalent to

$\displaystyle \frac{a^3+b^3+c^3+3abc}{2abc}\ge xy+yz+zx.$

Further,

\displaystyle\begin{align} &\frac{a^3+b^3+c^3+3abc}{2abc}\ge xy+yz+zx&\Leftrightarrow\\ &\frac{a^3+b^3+c^3+3abc}{2abc}\ge\frac{4}{(a+b)(b+c)(c+a)}\cdot [ab(a+b)+bc(b+c)+ca(c+a)]. \end{align}

By the AM-GM inequality, $\displaystyle \frac{1}{2abc}\ge\frac{4}{(a+b)(b+c)(c+a)},$ implying

\displaystyle\begin{align}&\frac{1}{2abc}\cdot [ab(a+b)+bc(b+c)+ca(c+a)]\\ &\qquad\ge\frac{4}{(a+b)(b+c)(c+a)}\cdot [ab(a+b)+bc(b+c)+ca(c+a)].\end{align}

Thus, suffice it to show that

$\displaystyle \frac{a^3+b^3+c^3+3abc}{2abc}\ge \frac{1}{2abc}\cdot [ab(a+b)+bc(b+c)+ca(c+a)],$

which is equivalent to

$a^3+b^3+c^3+3abc\ge ab(a+b)+bc(b+c)+ca(c+a),$

which is Schur's inequality of degree 3.

Solution 2

\displaystyle \begin{align} &a^3+b^3+c^3\geq \frac{(a+b+c)^3}{9}~\text{(Jensen's)} \\ &\frac{16}{(a+b)(b+c)(c+a)}\geq \frac{54}{(a+b+c)^3}~\text{(AM-GM)}. \end{align}

$\displaystyle x~\overset{\text{def}}{=} ~\frac{(a+b+c)^3}{9} \geq 3abc = 3~\text{(AM-GM)}.$

From the above three inequalities,

\displaystyle\begin{align} a^3+b^3+c^3+\frac{16}{(a+b)(b+c)(c+a)}-5&\geq x+\frac{6}{x}-5\\ &=\frac{(x-2)(x-3)}{x}\geq 0. \end{align}

Acknowledgment

Leo Giugiuc has kindly emailed me a problem of his, along with his own solution.

Solution 2 is by Amit Itagi.

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Copyright © 1996-2018 Alexander Bogomolny
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